Let the cost of a hat, shirt and jacket be H, S and J respectively.
3H+2S+J=140; 2H+2S+2J=170; 2J=180, so J=90; but something is wrong because this would make 2H+2S negative. Ignore the third equation and solve the first two in terms of J: 3H+2S=140-J and 2H+2S=170-2J. Subtract the latter from the former: H=-30+J or H=J-30. Put this value of H into H+S=85-J (dividing the equation by 2) and we get J-30+S=85-J; S=115-2J. The right-hand side must be positive so 115-2J>0 and J<115/2=57.50, so the jacket must be cheaper than $57.50. Also J-30>0 so J>30, the jacket is dearer than $30. Now we know that the jacket cost between $30 and $57.50. So 2J is between $60 and $115. H<27.50 and S<55 because of the limits on J. The third equation includes 2J and perhaps two more items adding up to $180. So the sum of up to two more items is between $120 and $65, making the average price between $60 and $32.50.
A graph helps, but doesn't solve the problem. Three equations can be plotted on the same graph: H=J-30; S=115-2J and Y=180-2J, where Y is the sum of other items that, together with the cost of two jackets, comes to $180. J is the "x" variable on the horizontal axis on which H, S and Y depend. The graph shows a number of interesting things.
The line S=115-2J crosses H=J-30 when J=$48.33 and H=S=$18.33. To the left of the intersection the hat is cheaper than the T-shirt; to the right the hat is dearer. Let's assume the hat is more expensive than the T-shirt. So the jacket costs between $48.33 and $57.50, the hat costs between $18.33 and $27.50, and the T-shirt less than $18.33.
The line Y=180-2J is parallel to S=115-2J and Y is between $65 and $83.33.
If the hat costs $20, then the jacket costs $50 and the T-shirt $15, these values fit the first two equations. What about the third? Y=$80=4S+H (for example). If the third statement had been "Four shirts, a hat and two jackets cost $180", then the answer would have been: jacket costs $50, hat costs $20 and the T-shirt costs $15.